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(06.03 MC)

An archer shoots an arrow up towards a target located on a hill, which is shown by the graph.

A graph of a line and a parabola intersecting at (48.63, 10.7).

Which set of equations best models the point of intersection of the arrow and the target?

y = –0.004x2 + 0.25x + 8 and y = x
y = 0.004x2 + 0.25x + 8 and y = 0.22x
y = –0.004x2 + 0.25x + 8 and y = 0.22x
y = –0.004x2 + 0.25x + 8 and y = –x


Sagot :

The set of equation that models the point of intersection of the arrow

and the target is the option; y = -0.004·x² + 0.25·x + 8 and y = 0.22·x

How can the given equations be analyzed?

The given parameters are;

Point of intersection of the line and the parabola is the point (48.63, 10.7)

Therefore, the point (48.63, 10.7) satisfies both the equation of the

parabola and the equation of the line, which by plugging in the value x =

48.67 in the different parabolas, we have;

The first set;

  • y = -0.004 × 48.67² + 0.25 × 48.67 + 8 ≈ 10.7, which corresponds with the ordered pair.

The linear function in the first set should be excluded, given y ≠ x in the

ordered pair.

In third set, y = -0.004 × 48.67² + 0.25 × 48.67 + 8 ≈ 10.7, while, from the

ordered pair, we have;

y = C·x

10.7 = 48.63·x

[tex]C = \dfrac{10.7}{48.63} \approx \mathbf{0.22}[/tex]

Therefore;

y ≈ 0.22·x

The set of equations that best models the point of intersection of the arrow and target are therefore;

  • y = -0.004·x² + 0.25·x + 8 and y = 0.22·x

Learn more about the equation of a parabola here:

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